This page will document useful transformations of elliptic integrals and relations among them, along with their derivations. These can be applied to simplifications of analytic formulae or numerical evaluations.

Contents

Definitions of Integrals
Complete Integrals of Negative Parameter
Reciprocal Modulus Transformations
Elliptic Nome on the Real Axis
Relation of First Kind to Other Forms

Definitions of Integrals

The incomplete elliptic integrals of the first, second and third kinds in trigonometric form are

$F\left(\phi \right|m)=\underset{0}{\overset{\phi }{\int }}\frac{d\phi }{\sqrt{1-m{sin}^2\phi }}$

$E\left(\phi \right|m)=\underset{0}{\overset{\phi }{\int }}d\phi \sqrt{1-m{sin}^2\phi }$

$\mathrm{\Pi }(n;\phi |m)=\underset{0}{\overset{\phi }{\int }}\frac{d\phi }{(1-n{sin}^2\phi )\sqrt{1-m{sin}^2\phi }}$

with $0<m<1$ in these defining integrals. The elliptic parameter m will be used throughout this presentation rather than the traditional elliptic modulus k for two reasons:

1) The resulting formulae are generally cleaner and simpler.

2) Computer algebra systems generally use the parameter rather than the modulus. Having formulae in the same form reduces unintended errors in application.

The elliptic parameter and modulus are related by $m=k^2$. There is also traditionally a complementary elliptic modulus defined by ${k^{\prime }}^2=1-k^2$, and functions of this complementary modulus are often denoted with a prime.

While there is a variety of mathematical notation for the elliptic integrals, it appears to be common to distinguish those using the parameter with a vertical bar before that argument, while those using the modulus are denoted with a comma before that argument.

The transformation $x=sin\phi$ gives the incomplete elliptic integrals in Legendre normal form:

$F\left(\phi \right|m)=\underset{0}{\overset{sin\phi }{\int }}\frac{dx}{\sqrt{(1-x^2)(1-m{x}^2)}}$

$E\left(\phi \right|m)=\underset{0}{\overset{sin\phi }{\int }}dx\sqrt{\frac{1-m{x}^2}{1-x^2}}$

$\mathrm{\Pi }(n;\phi |m)=\underset{0}{\overset{sin\phi }{\int }}\frac{dx}{(1-n{x}^2)\sqrt{(1-x^2)(1-m{x}^2)}}$

The angular variable is left as an argument on the left-hand sides to aid in avoiding errors in application.

Setting the angular argument equal to $\frac{\pi }{2}$ defines the complete elliptic integrals:

$K\left(m\right)=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{\sqrt{1-m{sin}^2\phi }}$

$E\left(m\right)=\underset{0}{\overset{\frac{\pi }{2}}{\int }}d\phi \sqrt{1-m{sin}^2\phi }$

$\mathrm{\Pi }\left(n\right|m)=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{(1-n{sin}^2\phi )\sqrt{1-m{sin}^2\phi }}$

Complete Integrals of Negative Parameter

A simple angular substitution of the form

$\begin{array}{llll}\phi =\frac{\pi }{2}-\theta & & & sin\phi =cos\theta \\ d\phi =-d\theta & & & cos\phi =sin\theta \end{array}$

interchanges sines and cosines while reversing the order of integration. This implies that the complete integrals can be equally defined using sines or cosines, since the integrals are otherwise identical after this substitution.

That fact can be used to evaluate the complete integrals for negative parameter quite easily:

$\begin{array}{l}K(-m)=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{\sqrt{1+m{sin}^2\phi }}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{\sqrt{m+1-m{cos}^2\phi }}\\ K(-m)=\frac{1}{\sqrt{m+1}}K\left(\frac{m}{m+1}\right)\end{array}$

$\begin{array}{l}E(-m)=\underset{0}{\overset{\frac{\pi }{2}}{\int }}d\phi \sqrt{1+m{sin}^2\phi }=\underset{0}{\overset{\frac{\pi }{2}}{\int }}d\phi \sqrt{m+1-m{cos}^2\phi }\\ E(-m)=\sqrt{m+1}E\left(\frac{m}{m+1}\right)\end{array}$

$\begin{array}{l}\mathrm{\Pi }\left(n\right|-m)=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{(1-n{sin}^2\phi )\sqrt{1+m{sin}^2\phi }}\\ \phantom{\mathrm{\Pi }\left(n\right|-m)}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{(1-n+n{cos}^2\phi )\sqrt{m+1-m{cos}^2\phi }}\\ \mathrm{\Pi }\left(n\right|-m)=\frac{1}{(1-n)\sqrt{m+1}}\mathrm{\Pi }\left(\frac{n}{n-1}\right|\frac{m}{m+1})\end{array}$

Reciprocal Modulus Transformations

With an argument of $\frac{1}{m}$ the denominator of the complete integral of the first kind acquires a zero. Since the real part of the inverse sine function is always less than $\frac{\pi }{2}$ , split the integral into two parts:

$K\left(\frac{1}{m}\right)=\underset{0}{\overset{{sin}^{-1}\sqrt{m}}{\int }}\frac{d\phi }{\sqrt{1-\frac{1}{m}{sin}^2\phi }}-i\underset{{sin}^{-1}\sqrt{m}}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{\sqrt{\frac{1}{m}{sin}^2\phi -1}}$

These two integrals can be converted to complete elliptic integrals with trigonometric substitutions that preserve the valid endpoint while scaling the other to a value appropriate for a complete integral. For the integral in the real part, the substitution that preserves the lower endpoint is

$sin\phi =\sqrt{m}sin\alpha cos\phi d\phi =\sqrt{m}cos\alpha d\alpha$

under which the integral becomes

$\underset{0}{\overset{{sin}^{-1}\sqrt{m}}{\int }}\frac{d\phi }{\sqrt{1-\frac{1}{m}{sin}^2\phi }}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sqrt{m}cos\alpha d\alpha }{\sqrt{1-{sin}^2\alpha }\sqrt{1-m{sin}^2\alpha }}=\sqrt{m}K\left(m\right)$

For the integral in the imaginary part, the substitution that preserves the upper endpoint is

$cos\phi =\sqrt{1-m}cos\alpha sin\phi d\phi =\sqrt{1-m}sin\alpha d\alpha$

under which the integral becomes

$\underset{{sin}^{-1}\sqrt{m}}{\overset{\frac{\pi }{2}}{\int }}\frac{d\phi }{\sqrt{\frac{1}{m}{sin}^2\phi -1}}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sqrt{1-m}sin\alpha d\alpha }{\sqrt{\frac{1-(1-m){cos}^2\alpha }{m}-1}\sqrt{1-(1-m){cos}^2\alpha }}=\sqrt{m}K(1-m)$

since the complete integrals can be defined with either sines or cosines as noted above. The reciprocal modulus transformation for the complete elliptic integral of the first kind is thus

$K\left(\frac{1}{m}\right)=\sqrt{m}\left[K\right(m)-iK(1-m\left)\right]$

This result technically only holds for $Imm>0$ : for $Imm<0$ the imaginary part of the result requires a change in sign.

For $m>1$ this relation can be used to separate the real and imaginary parts of the complete integral when written in the form

$K\left(m\right)=\frac{1}{\sqrt{m}}\left[K\right(\frac{1}{m})-iK(1-\frac{1}{m}\left)\right]$

since then the arguments on the right-hand side are both within the defining range of convergence.

Elliptic Nome on the Real Axis

The elliptic nome is defined by

$q\left(m\right)=exp[-\pi \frac{K(1-m)}{K\left(m\right)}]$

For $0<m<1$ the arguments of the elliptic integrals are within the defining range, so that both integrals as well as the nome are real.

For $m<0$ the arguments of both integrals are outside the defining range. The complete integral with negative parameter has been given above, and the integral with argument greater than one can be rewritten using the reciprocal modulus transformation. The ratio of the two integrals is

$\frac{K(1+m)}{K(-m)}=\frac{\sqrt{m+1}}{K\left(\frac{m}{m+1}\right)}\times \frac{1}{\sqrt{m+1}}\left[K\right(\frac{1}{m+1})-iK(1-\frac{1}{m+1}\left)\right]=\frac{K\left(\frac{1}{m+1}\right)}{K\left(\frac{m}{m+1}\right)}-i$

so that the elliptic nome for real negative argument is

$q(-m)=-exp[-\pi \frac{K\left(\frac{1}{m+1}\right)}{K\left(\frac{m}{m+1}\right)}]$

and since both arguments are now within the defining range, the nome is purely real.

For $m>1$ the arguments of both integrals are again outside the defining range but with their respective roles reversed. The ratio of the two integrals is now

$\begin{array}{l}\frac{K(1-m)}{K\left(m\right)}=\frac{K[-(m-1\left)\right]}{K\left(m\right)}\\ \phantom{\frac{K(1-m)}{K\left(m\right)}}=\frac{K\left(\frac{m-1}{m}\right)}{\sqrt{m}}\times \frac{\sqrt{m}}{K\left(\frac{1}{m}\right)-iK(1-\frac{1}{m})}\\ \frac{K(1-m)}{K\left(m\right)}=\frac{K(1-\frac{1}{m})}{K(\frac{1}{m}{)}^2+K(1-\frac{1}{m}{)}^2}\left[K\right(\frac{1}{m})+iK(1-\frac{1}{m}\left)\right]\end{array}$

The arguments on the right-hand side are now within the defining range, so that this ratio is explicitly separated into real and imaginary parts. The nome itself

$q\left(m\right)=exp\left[\pi \frac{K(1-\frac{1}{m})}{K(\frac{1}{m}{)}^2+K(1-\frac{1}{m}{)}^2}\left[K\right(\frac{1}{m})+iK(1-\frac{1}{m}\left)\right]\right]$

is no longer entirely real in this domain.

Relation of First Kind to Other Forms

The inverse of the Weierstrass elliptic function can be defined as an elliptic integral:

${\wp }^{-1}(x;g_2,g_3)=\underset{x}{\overset{\infty }{\int }}\frac{dt}{\sqrt{4t^3-g_{2}t-g_3}}=\frac{1}{2}\underset{x}{\overset{\infty }{\int }}\frac{dt}{\sqrt{(t-e_1)(t-e_2)(t-e_3)}}$

With the change of variable

$\begin{array}{l}{u}^2=\frac{e_2-e_1}{t-e_1}\\ t=\frac{e_2-e_1}{u^2}+e_1\end{array}2udu=-\frac{e_2-e_1}{(t-e_1{)}^2}dt=-\frac{u^4}{e_2-e_1}dt$

the integral becomes

$\frac{1}{2}\underset{x}{\overset{\infty }{\int }}\frac{dt}{\sqrt{(t-e_1)(t-e_2)(t-e_3)}}=\frac{1}{\sqrt{e_2-e_1}}\underset{0}{\overset{\sqrt{\frac{e_2-e_1}{x-e_1}}}{\int }}\frac{du}{\sqrt{(1-u^2)(1-\frac{e_3-e_1}{e_2-e_1}{u}^2)}}$

The right-hand side is the Legendre normal form of the integral of the first kind. Keeping in mind the extra sine in its definition, one can write

${\wp }^{-1}(x;g_2,g_3)=\frac{1}{\sqrt{e_2-e_1}}F\left({sin}^{-1}\sqrt{\frac{e_2-e_1}{x-e_1}}\right|\frac{e_3-e_1}{e_2-e_1})$

where the Weierstrass roots e_i are evaluated from the Weierstrass invariants g_i . It may at first appear that the right-hand side is critically dependent on the ordering of the roots, but given the symmetry of the Weierstrass form this is not true. The change of variable can be written in six different ways, each of which will provide a valid result with an appropriate permutation of the roots.

Actual numerical implementation of this equivalence runs into problems with branch points. Here is a straightforward graphic comparing the inverse Weierstrass function in blue to the right-hand side in red:

While the relation is clearly valid for some extent of the domain for both the real and imaginary parts, in other regions there are apparent constant displacements and inversions. This graphic acts as a test of the library Math and will be left as is as a reminder of the difficulties of handling branch points numerically.

The relation between the two elliptic integrals can be written with the independent argument on the right-hand side, but the resultant mixture of Weierstrass roots and invariants on the left-hand side would be a bit confusing.

With an offset in the variable of integration in the definition of the inverse Weierstrass function, along with a definition of a Carlson symmetric integral, one can write

${\wp }^{-1}(x;g_2,g_3)=\frac{1}{2}\underset{0}{\overset{\infty }{\int }}\frac{dt}{\sqrt{(t+x-e_1)(t+x-e_2)(t+x-e_3)}}=R_F(x-e_1,x-e_2,x-e_3)$

which gives the immediate relation

$R_F(x-e_1,x-e_2,x-e_3)=\frac{1}{\sqrt{e_2-e_1}}F\left({sin}^{-1}\sqrt{\frac{e_2-e_1}{x-e_1}}\right|\frac{e_3-e_1}{e_2-e_1})$

The relation between the inverse Weierstrass function and the Carlson symmetric integral is used to evaluate the former in Math.

Uploaded 2017.11.20 — Updated 2020.08.15 analyticphysics.com